Question #a0129

I got `"0.4072 atm"`.

This implies that by knowing

`"0.1026 atm" harr 20.00^@ "C"`
`"0.4072 atm" harr 54.00^@ "C"`
`"1.000 atm" harr 80.74^@ "C"`

the slope of the liquid-vapor coexistence curve on the phase diagram of cyclohexane is positive. That's normal, as most if not all liquids expand when they vaporize...

The Clapeyron equation approximates the slope of a two-phase equilibrium line:

`(dP)/(dT) = (DeltabarH)/(TDeltabarV)`

where:

• `DeltabarH` is the change in molar enthalpy.
• `T` is the temperature in `"K"`.
• `barV = V/n` is the molar volume of the substance.

In considering boiling point we examine the liquid-vapor coexistence curve (`bar(TC)`).

So, assuming the vapor is ideal, `barV = (RT)/P`, and:

`(dP)/(dT) = (DeltabarH cdot P)/(RT^2)`

Therefore, we can then write the derivative form of the Clausius-Clapeyron equation:

`1/P(dP)/(dT) = (DeltabarH_"vap")/(RT^2) = (dlnP)/(dT)`

You may recognize this better once we separate the variables and integrate.

`int_(P_1)^(P_2)1/PdP = int_(T_1)^(T_2)(DeltabarH_"vap")/(RT^2)dT`

We assume the change in molar enthalpy due to vaporization remains constant in the small temperature range (i.e. that this curve is straight!) to obtain the Clausius-Clapeyron Equation:

`=> color(green)barul(|stackrel(" ")(" "ln(P_2/P_1) = -(DeltabarH_"vap")/R [1/T_2 - 1/T_1]" ")|)`

and if you were to plot the `ln` of a set of pressure data points on the `y` axis and a set of corresponding inverse boiling points `1/T` on the `x` axis, you would obtain a slope of `-(DeltabarH_"vap")/R`.

And now our equation relates vapor pressures `P_i` with boiling temperatures `T_i`, using `DeltabarH_"vap"` as part of our slope.

Now, you were given:

• The normal boiling point of `80.74^@ "C"` occurs implicitly at `"1.000 atm"` vapor pressure.
• At `20.00^@ "C"`, the vapor pressure is `"0.1026 atm"`.

At this point you can then solve for the enthalpy of vaporization.

`ln("1.000 atm"/"0.1026 atm") = -(DeltabarH_"vap")/("0.008314472 kJ/mol"cdot"K") [1/(80.74 + "273.15 K") - 1/(20.00 + "273.15 K")]`

Therefore:

`DeltabarH_"vap" = -ln(9.747) cdot "0.008314472 kJ/mol"cdot"K" cdot [-5.855 xx 10^(-4) "K"^(-1)]^(-1)`

`=` `ul"32.33 kJ/mol"`

Apparently the actual value is around `"30.1 kJ/mol"`, so we're not that far off.

So now, if we want the vapor pressure at `54.00^@ "C"`, we can use the `20.00^@ "C"` as a new `T_1` reference point and solve for `P_2` using `T_2 = 54.00^@ "C"`. That is, we found what we needed to get to the slope, and now we are re-using it to find a single data point.

`ln(P_2/"0.1026 atm") = -("32.33 kJ/mol")/("0.008314472 kJ/mol"cdot"K") [1/(54.00 + "273.15 K") - 1/(20.00+"273.15 K")]`

`= 1.379`

As a result, the vapor pressure at the intermediate temperature is:

`color(blue)(P_2) = "0.1026 atm" cdot e^(1.379) ~~ color(blue)ul("0.4072 atm")`